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Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$. Suppose the characteristic polynomial of $T$ is $f(t) = (t-\lambda)^3$ for the eigenvalue $\lambda$. Suppose also that $\dim E_\lambda = 2$, and that $\beta = \{v_1,v_2\}$ is a basis for this eigenspace.

If I wanted to create the Jordan Canonical Form Matrix for $T$, I would require a basis $\alpha = \beta \cup \text{span}(v_3)$ where $v_3 \in K_\lambda$, the generalized Eigenspace of $\lambda$.

Let $a,b$ be vectors such that

$$(T-\lambda I)a = v_1 \qquad (T-\lambda I)b = v_2$$

Would both $a$ and $b$ be appropriate choices for $v_3$?

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Either is an appropriate choice assuming it exists. But at most one of $a$ and $b$ can exist, and it's possible that neither exists.

You have to choose $v_1$ and $v_2$ more carefully. You should really find $a$ or $b$ first, i.e., find a solution to the set of equations $$(T-\lambda I)^2 a = 0 \\ (T-\lambda I) a \ne 0$$ and then set $v_1 = (T-\lambda I)a$. Then find another $\lambda$-eigenvector $v_2$ which is independent from $v_1$.

Also, if the characteristic polynomial is $(t-\lambda)^3$, then the generalized $\lambda$-eigenspace is all of $\mathbb{R}^3$. So you need a better $v_3$ than you claim you do.

At most one can exist:

If both existed, then $\{v_1,v_2,a,b\}$ would be linearly independent: if $$c_1v_1 + c_2a + d_1v_2 + d_2b = 0,$$ with $c_1,c_2,d_1,d_2 \in \mathbb{R}$, then apply $(T-\lambda I)$ to both sides, so that $$c_2v_1 + d_2v_2 = 0,$$ which implies $c_2=d_2=0$ since $\{v_1,v_2\}$ is a linearly independent set, and then $$c_1v_1 + d_1v_2 = 0,$$ and so $c_1=d_1=0$ for the same reason. But $\dim(\mathbb{R}^3) < 4$.

It's possible that neither exists:

Consider $$T=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix},$$ which has characteristic polynomial $(t-1)^3$. Taking $$v_1 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \text{ and }\; v_2= \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix},$$ the equations $(T-I)a = v_1$ and $(T-I)b = v_2$ have no solutions, but $\{v_1,v_2\}$ is a basis for the $1$-eigenspace.

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  • $\begingroup$ In your example where neither exists, it is guaranteed that there will exist another set of vectors, say $w_1,w_2$ that would form a basis such that one of $a,b$ will exist? $\endgroup$
    – Bryden C
    Apr 3, 2019 at 17:26
  • $\begingroup$ Yes, because the eigenvalues are real (not complex) and it can be proven that any such matrix can be put into Jordan normal form over $\mathbb{R}$. Take $w_1 = (1,0,0)^T$ and $w_2 = (0,1,0)^T$. Then $b = (0,0,1)^T$ is a solution to $(T-I)b = w_2$. (Hint: I put $T$ in JNF to begin with.) $\endgroup$
    – user555203
    Apr 3, 2019 at 17:28

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